If the length of the perpendicular drawn from | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The equation of the line with intercepts $a$ and $b$ is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $\frac{1}{a}x + \frac{1}{b}y - 1 =

Vedclass · Accepted Answer

$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$

Vedclass · Answer

(D) The equation of the line with intercepts $a$ and $b$ is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $\frac{1}{a}x + \frac{1}{b}y - 1 = 0$.The length of the perpendicular $p$ from the origin $(0, 0)$ to this line is given by $p = \frac{|\frac{1}{a}(0) + \frac{1}{b}(0) - 1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}}$.This simplifies to $p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.Squaring both sides,we get $p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}}$.Therefore,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$.

If the length of the perpendicular drawn from the origin to the line whose intercepts on the axes are $a$ and $b$ is $p$,then

Similar Questions

Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$. (in $/5$)

If $O$ is the origin and $A$ and $B$ are points on the line $3x - 4y + 25 = 0$ such that $OA = OB = 13$,then the area of $\triangle OAB$ (in sq units) is

Let $\alpha$ be the distance between the lines $-x + y = 2$ and $x - y = 2$,and $\beta$ be the distance between the lines $4x - 3y = 5$ and $6y - 8x = 1$,then:

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to line $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is $\qquad$ units.

If $p$ and $q$ are the perpendicular distances from the origin to the straight lines $x \sec \theta - y \operatorname{cosec} \theta = a$ and $x \cos \theta + y \sin \theta = a \cos 2 \theta$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module